## 17.18 Duals

Let $(X, \mathcal{O}_ X)$ be a ringed space. The category of $\mathcal{O}_ X$-modules endowed with the tensor product constructed in Section 17.16 is a symmetric monoidal category. For an $\mathcal{O}_ X$-module $\mathcal{F}$ the following are equivalent

1. $\mathcal{F}$ has a left dual in the monoidal category of $\mathcal{O}_ X$-modules,

2. $\mathcal{F}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module, and

3. $\mathcal{F}$ is of finite presentation and flat as an $\mathcal{O}_ X$-module.

This is proved in Example 17.18.1 and Lemmas 17.18.2 and 17.18.3 of this section.

Example 17.18.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module which is locally a direct summand of a finite free $\mathcal{O}_ X$-module. Then the map

$\mathcal{F} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F})$

is an isomorphism. Namely, this is a local question, it is true if $\mathcal{F}$ is finite free, and it holds for any summand of a module for which it is true. Denote

$\eta : \mathcal{O}_ X \longrightarrow \mathcal{F} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X)$

the map sending $1$ to the section corresponding to $\text{id}_\mathcal {F}$ under the isomorphism above. Denote

$\epsilon : \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \otimes _{\mathcal{O}_ X} \mathcal{F} \longrightarrow \mathcal{O}_ X$

the evaluation map. Then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X), \eta , \epsilon$ is a left dual for $\mathcal{F}$ as in Categories, Definition 4.43.5. We omit the verification that $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_\mathcal {F}$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X)}$.

Lemma 17.18.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Let $\mathcal{G}, \eta , \epsilon$ be a left dual of $\mathcal{F}$ in the monoidal category of $\mathcal{O}_ X$-modules, see Categories, Definition 4.43.5. Then

1. $\mathcal{F}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module,

2. the map $e : \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{O}_ X) \to \mathcal{G}$ sending a local section $\lambda$ to $(\lambda \otimes 1)(\eta )$ is an isomorphism,

3. we have $\epsilon (f, g) = e^{-1}(g)(f)$ for local sections $f$ and $g$ of $\mathcal{F}$ and $\mathcal{G}$.

Proof. The assumptions mean that

$\mathcal{F} \xrightarrow {\eta \otimes 1} \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \xrightarrow {1 \otimes \epsilon } \mathcal{F} \quad \text{and}\quad \mathcal{G} \xrightarrow {1 \otimes \eta } \mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} \xrightarrow {\epsilon \otimes 1} \mathcal{G}$

are the identity map. Let $x \in X$. We can find an open neighbourhood $U$ of $x$, a finite number of sections $f_1, \ldots , f_ n$ and $g_1, \ldots , g_ n$ of $\mathcal{F}$ and $\mathcal{G}$ over $U$ such that $\eta (1) = \sum f_ i g_ i$. Denote

$\mathcal{O}_ U^{\oplus n} \to \mathcal{F}|_ U$

the map sending the $i$th basis vector to $f_ i$. Then we can factor the map $\eta |_ U$ over a map $\tilde\eta : \mathcal{O}_ U \to \mathcal{O}_ U^{\oplus n} \otimes _{\mathcal{O}_ U} \mathcal{G}|_ U$. We obtain a commutative diagram

$\xymatrix{ \mathcal{F}|_ U \ar[rr]_-{\eta \otimes 1} \ar[rrd]_-{\tilde\eta \otimes 1} & & \mathcal{F}|_ U \otimes \mathcal{G}|_ U \otimes \mathcal{F}|_ U \ar[r]_-{1 \otimes \epsilon } & \mathcal{F}|_ U \\ & & \mathcal{O}_ U^{\oplus n} \otimes \mathcal{G}|_ U \otimes \mathcal{F}|_ U \ar[u] \ar[r]^-{1 \otimes \epsilon } & \mathcal{O}_ U^{\oplus n} \ar[u] }$

This shows that the identity on $\mathcal{F}$ locally on $X$ factors through a finite free module. This proves (1). Part (2) follows from Categories, Lemma 4.43.6 and its proof. Part (3) follows from the first equality of the proof. You can also deduce (2) and (3) from the uniqueness of left duals (Categories, Remark 4.43.7) and the construction of the left dual in Example 17.18.1. $\square$

Lemma 17.18.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a flat $\mathcal{O}_ X$-module of finite presentation. Then $\mathcal{F}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module.

Proof. After replacing $X$ by the members of an open covering, we may assume there exists a presentation

$\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{F} \to 0$

Let $x \in X$. By Lemma 17.17.10 we can, after shrinking $X$ to an open neighbourhood of $x$, assume there exists a factorization

$\mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1} \to \mathcal{F}$

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1}$ annihilates the first summand of $\mathcal{O}_ X^{\oplus r}$. Repeating this argument $r - 1$ more times we obtain a factorization

$\mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F}$

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r}$ is zero. This means that the surjection $\mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F}$ has a section and we win. $\square$

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